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Let $(M, \omega)$ be a symplectic 4-manifold. An immersed symplectic surface is an immersion $v: S \to M$ such that $v^*\omega>0$ is a volume form.

Fix a tame/compatible almost complex structure $J$. A holomorphic curve is a map $u:(\Sigma, j) \to (M, J)$ such that $$du+J\circ du \circ j=0.$$

If $u$ is an immersion holomorphic curve, then it is also an immersed symplectic surface. The reason is that $u^*\omega(\partial_x, \partial_y)=\omega(\partial_xu, \partial_yu)= \omega(\partial_xu, J\partial_xu)>0$, where $(x, y)$ are holomorphic coordinates on $\Sigma.$

Conversely, I want to ask given an immersed symplectic surface $v: S \to M$, can we find a tame/compatible almost complex structure $J$ such that $v$ is holomorphic?

seumath12345
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  • Have you attempted the embedded case? – J.V.Gaiter Jan 14 '24 at 22:13
  • In the embedded case, it seems to be true. Because a neighborhood of the symplectic surface can be identify with its normal bundle symplectically, we can construct almost complex structure in the neighborhood such that $S$ is holomorphic, and then extend it to the whole manifold. But I don't know whether this can be generalized to the immersed case. Say $S$ has double points, we cannot just identify the neighborhood with the normal bundle? – seumath12345 Jan 15 '24 at 00:06

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